Example of algebraic field not honoring the Total Order relation?

I've started studying Calculus. I've stumbled upon the definition of an 'Ordered field'. One of the requirements are for the field to honor the Total Order relation. Meaning that for every a,b, a could either be:

• a > b
• a < b
• a = b

While this sounds trivial, I'm trying for intuition purposes to find an counter example of a field not honoring the Total Order relation. Can you give me an example?

Thanks

-

$\Bbb C$, the field of complex numbers. If $\Bbb C$ were an ordered field, ordered by some linear order $\le$, the set $P=\{z\in\Bbb C:z>0\}$ would have the following properties:

1. For every $z\in\Bbb C$, exactly one of the following is true: $z\in P$, $-z\in P$, $z=0$.
2. $P$ is closed under multiplication: if $w,z\in P$, then $wz\in P$.

Now $i\ne 0$, so by (1) either $i\in P$ or $-i\in P$. If $i\in P$, then $-1=i^2\in P$ by (2). If, on the other hand, $-i\in P$, then $-1=(-i)^2\in P$. Thus, we can be sure that $-1\in P$. But then $1=(-1)^2\in P$, so both $-1$ and $1$ are in $P$, violating (1). Thus, there is no linear order that makes $\Bbb C$ an ordered field.

-
ah thanks! wasn't thinking of imaginary numbers. –  vondip Nov 5 '12 at 7:53
@vondip: You’re welcome! –  Brian M. Scott Nov 5 '12 at 7:55

It is rather immediate that if a field $K$ admits a total ordering, then $-1$ is not a sum of squares.

Indeed, if we have a total ordering on $K$, then $1>0$, thus $-1<0$. Also for every $a\in K$ we have $a^2>0$. Thus $-1\neq \sum_{i=1}^n a_i^2$, since one side is positive and the other is negative.

From this observation one immediately concludes that fields of positive characteristic (e.g. finite fields) do not admit total orderings. (A field $K$ has positive characteristic if there exists $n>0$ such that $\underbrace{1+\cdots+1}_{n\ \textrm{times}}=0$ in $K$.)

It also gives a way to construct fields, say, sub-fields of $\mathbb{C}$ that admit no total ordering.

For example the cyclotomic fields: Let $n>2$ and let $\zeta = e^{2\pi i/n}$ be a primitive $n$-th root of unity. Then $\mathbb{Q}(\zeta)$, the minimal sub-field of $\mathbb{C}$ that contains $\zeta$, admits no total ordering. To see this we separate between two cases. In the first case $n$ is even, and then $-1=\zeta^{n/2}$ is a square.

In the second case $n$ is odd. Then the sets $\{\zeta^{k} \mid k=0, \ldots, n-1\}$ and $\{\zeta^{2k} \mid k=0, \ldots,n-1\}$ coincide. So using the formula for the sum of terms in geometric series we get that

$\displaystyle\zeta^{n-1} + \zeta^{n-2} + \cdots + \zeta +1 = \frac{1-\zeta^n}{1-\zeta} =0,$

so

$\displaystyle -1=\zeta^{n-1} + \zeta^{n-2} + \cdots + \zeta = \zeta^{2(n-1)} + \zeta^{2(n-2)} + \cdots + \zeta^2 = (\zeta^{n-1})^2 + (\zeta^{n-2})^2 + \cdots + \zeta^2$

is a sum of squares.

-

In fact, a field $k$ can be ordered iff $-1$ is not a sum of squares.

If $-1$ is a sum of squares in an ordered field, then $-1$ is positive whence $1$ is negative; but $1$ is a square and so positive. It is a contradiction.

Let us suppose that $-1$ is not a sum of squares. Then set of field $K \supset k$ in which $-1$ is not a sum of squares is inductive; by Zorn's lemma, there exists a maximal field $\tilde{k}$ containing $k$ in which $-1$ is not a sum of squares.

Let $\gamma \in \tilde{k}$ not be a square. Then $\tilde{k}\subsetneq \tilde{k}(\sqrt{\gamma})$ where $\tilde{k}(\sqrt{\gamma})$ is a rupture field of $X^2-\gamma$. By definition of $\tilde{k}$, there exist $a_i$ and $b_i$ such that $-1= \sum\limits_{i=1}^n (a_i+ \sqrt{\gamma}b_i)= \sum\limits_{i=1}^n (a_i^2+ \gamma b_i^2) + 2 \sqrt{\gamma}\sum\limits_{i=1}^n a_ib_i$. Because $\sqrt{\gamma} \notin \tilde{k}$, $\sum\limits_{i=1}^na_ib_i=0$ and because $-1$ is not a sum of squares in $\tilde{k}$, $\gamma$ is not a sum of squares in $\tilde{k}$.

So, in $\tilde{k}$, a sum of squares is a square. Because $-1$ is not a sum of squares in $\tilde{k}$, $\sum\limits_{i=1}^n b_i^2 \neq 0$. Then we deduce that $\displaystyle - \gamma = \frac{1^2+ \sum\limits_{i=1}^n a_i^2}{\sum\limits_{i=1}^n b_i^2}$ is a square. So, for all $\gamma \in \tilde{k}$, $\gamma$ or $-\gamma$ is a square. Finally, define on $k$ the following order: $y \leq x$ iff $x-y$ is a square in $\tilde{k}$ and prove that $(k,\leq)$ is an ordered field.

For example, $\mathbb{C}$ and $\mathbb{F}_{p^n}$ cannot be ordered because $-1=i^2$ in $\mathbb{C}$ and $-1= \underset{p^n-1}{\underbrace{1^2+...+1^2}}$ in $\mathbb{F}_{p^n}$.

Reference: Algebra, van der Waerden.

-